This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 6 years ago.
I work on an Android project, in which I use AsyncTask to send nameValuePairs to webservice.
I use PHP code below to send data from application to database:
<?php
require_once('dbConnect.php');
$studentid = $_POST['name'];
$classid = $_POST['classid']
$studentid = $_POST['studentid'];
$start = $_POST['start'];
$end = $_POST['end'];
$startsig = $_POST['startsig'];
$endsig = $_POST['endsig'];
$sql = "insert into signature (studentid,classid,start,end,startsig,endsig) values ('$studentid','$classid','$start','$end','$startsig','$endsig')";
if(mysqli_query($con,$sql)){
echo 'success';
}
else{
echo 'failure';
}
mysqli_close($con);
?>
Values in dbConnect.php file are correct. Database schema looks like this.
I checked, application sends out data successfully, but when I try to access PHP file above from browser I get a HTTP 500 error.
What else should I check in this case?
There are two problems I noticed in your code you didn't put a semicolon after
$classid = $_POST['classid']
And also why are you assigning two different post values to same phpvariable $studentid = $_POST['name']; and $studentid = $_POST['studentid'];.Anyways $studentid will have the last assigned value $_POST['studentid']
Try debugging for more errors
Related
This question already has answers here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Reference - What does this error mean in PHP?
(38 answers)
Closed 3 years ago.
I can't insert the registration form data into my localhost database.
I am using xampp server, running mysql 5, php 7 and apache2. I've provided the code that can insert the form data into the database. The database is connected. I have also restarted the xampp server but it shows the same problem.
<?php
$active='Account';
include("includes/header.php");
?>
//html form with correct name values
<?php
if(isset($_POST['register'])){
$c_name = $_POST['c_name'];
$c_email = $_POST['c_email'];
$c_pass = $_POST['c_pass'];
$c_country = $_POST['c_country'];
$c_city = $_POST['c_city'];
$c_contact = $_POST['c_contact'];
$c_address = $_POST['c_address'];
$c_image = $_FILES['c_image']['name'];
$c_image_tmp = $_FILES['c_image']['tmp_name'];
move_uploaded_file($c_image_tmp,"customer/customer_images/$c_image");
$insert_customer = "insert into customers (customer_name,customer_email,customer_pass,customer_country,customer_city,customer_contact,customer_address,customer_image,customer_ip) values ('$c_name','$c_email','$c_pass','$c_country','$c_city','$c_contact','$c_address','$c_image','$c_ip')";
$result = mysqli_query($con,$insert_customer);
$sel_cart = "select * from cart where ip_add='$c_ip'";
$run_cart = mysqli_query($con,$sel_cart);
$check_cart = mysqli_num_rows($run_cart);
if($check_cart>0){
$_SESSION['customer_email']=$c_email;
echo "<script>alert('User is already present');</script>";
echo "<script>window.open('checkout.php','_self')</script>";
}else{
$_SESSION['customer_email']=$c_email;
echo "<script>alert('You have been Registered Sucessfully')</script>";
echo "<script>window.open('index.php','_self')</script>";
}
}
?>
The data should be inserted into the db when I refresh the phpmyadmin page.
Check $con variable where you define. Connection variable should be define on page and call with mysqli function.
$result = mysqli_query($con,$insert_customer);
This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
How can I prevent SQL injection in PHP?
(27 answers)
Closed 4 years ago.
Having trouble getting the data to insert into Maria DB. While attempting this project I have used about every source I can think of and I'm resorting to posting a question online. Below is my code and the error it produces is "Fatal error: Call to undefined method mysqli::mysqli_prepare() in C:\xampp\htdocs\scripts\insert.php on line 12". Been mulling over this for a couple days and so far I have found nothing at all on youtube or stackoverflow.
<?php
//Set up credentials
$servername = "localhost";
$username = "root";
$password = "";
$database = "test";
$db = new PDO("mysql:host=localhost;dbname=test;","$username","$password");
// Create connection
$link = new mysqli($servername,$username,$password,$database) or die($mysqli->error);
$a = $_POST["a"];
$sql = "INSERT INTO ass (asdf) VALUES('$a')";
$result = $link->mysqli_prepare($sql);
$stmt = $database->mysqli_prepare($sql);
$stmt->execute(array($_POST['a']));
$stmt->close();
?>
This question already has answers here:
How do I get PHP errors to display?
(27 answers)
Closed 5 years ago.
I am continuously having this internal server error to a very simple php code working with JSON. The values I am logging from the code is absolutely correct, yet when in action I am getting an internal server error 500 from this particular code. I had a similar code working previously. What am I doing wrong? or how should I proceed with debugging the error?
<?php
$var2 = $_POST['phn'];
$phone_received = json_decode($var2);
$adb = PearDatabase::getInstance();
$query1 = "SELECT addressid FROM address WHERE mobile = ?";
$leadID = $adb->pquery($query1, array($phone_received));
$row= $adb->num_rows($leadID);
if ($row != 0) {
$result = 'This number has already been used in the system.';
echo json_encode($result);
}else{
$result = 'Good to go!';
echo json_encode($result);
}
?>
Did you try to add true parameter to your json_decode call? I suppose the problem is with the $phone_received param. I also suggest you line by line debugging: try to retrun some dummy result after each line of code step by step so you can detect the place where the error occurs.
This question already has answers here:
How do I get PHP errors to display?
(27 answers)
Closed 6 years ago.
So, I have some bug in this code that will not execute the query, it just says The localhost page isn’t working - localhost is currently unable to handle this request. And that error shows only when there's some bug in the code, so it won't execute it. I don't know what's the problem here, as I tried to connect to database and that's all ok. Problem lies in lines underneath the connect.php.
if($_POST) {
include('connection.php');
$id_broj = $_POST['id_num'];
$password = $_POST['password'];
$query=mysqli_query("SELECT id,id_broj,ime,prezime FROM zaposleni");
$result=mysqli_query($conn,$query);
$count=mysqli_num_rows($result);
$row=mysqli_fetch_array($result);
echo 'Passed';
}
Need to write only query in $query Variable as:
if($_POST) {
include('connection.php');
$id_broj = $_POST['id_num'];
$password = $_POST['password'];
$query="SELECT id,id_broj,ime,prezime FROM zaposleni";
$result=mysqli_query($conn,$query);
$count=mysqli_num_rows($result);
$row=mysqli_fetch_array($result);
echo 'Passed';
}
This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 6 years ago.
Alright so I am currently trying to make a system to take applications and it was working until I added the rest of the questions :
<?php
include_once('db.php');
$user =$_POST['username'];
$job =$_POST['job'];
$active =$_POST['active'];
$why =$_POST['Q1']
$le =$_POST['Q2']
$skype =$_POST['skype']
if(mysqli_query($conn, "INSERT INTO app (username,job,active,why,le,skype) VALUES ('$user','$job','$active','$why','$le','$skype')"))
echo"successfully inserted";
else
echo "failed";
?>
But when I have it like that I get this error
Parse error: syntax error, unexpected '$le' (T_VARIABLE) in C:\xampp\htdocs\app_insert.php on line 8
Keep in mind I am using xampp with Apache and Mysql, anyone know whats happening?
You missed out the semi-colon, ; for:
$why =$_POST['Q1']
$le =$_POST['Q2']
$skype =$_POST['skype']
It should be:
$why =$_POST['Q1'];
$le =$_POST['Q2'];
$skype =$_POST['skype'];
You are missing the semi-colon and the insert SQL statement seems incorrect. You are passing the string to SQL statements instead of the value
<?php
include_once('db.php');
$user =$_POST['username'];
$job =$_POST['job'];
$active =$_POST['active'];
$why =$_POST['Q1'];
$le =$_POST['Q2'];
$skype =$_POST['skype']
if(mysqli_query($conn, "INSERT INTO app (username,job,active,why,le,skype) VALUES ('{$user}','{$job}','{$active}','{$why}','{$le}','{$skype}')"))
echo"successfully inserted";
else
echo "failed";
?>